Problem 20 - Introductory Algebra for College Students

Question

Solve each quadratic equation by the square root property. If possible, simplify radicals or rationalize denominators. $$(x-2)^{2}=25$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation \((x-2)^2 = 25\) are \(x=7\) and \(x=-3\).

Step 1: Apply the Square Root Property

To apply the square root property, one needs to take the square root on both sides of the equation \((x-2)^2 = 25\). It's important to remember that taking the square root yields two solutions, positive and negative. Therefore, the equation can be rewritten as \(x-2 = \pm \sqrt{25}\).

Step 2: Simplify the Square Root

The square root of 25 is 5, so the equation now becomes \(x - 2 = \pm 5\).

Step 3: Solve for x

Finally, to isolate \(x\) one adds 2 to both sides of the equation. This gives \(x = 2 \pm 5\). This equation actually represents two separate equations: one with the positive square root and one with the negative. So the solutions are \(x = 7\) (from the equation \(x = 2 + 5\)) and \(x = -3\) (from the equation \(x = 2 - 5\)).

Key Concepts

Square Root Property

The square root property is a valuable tool when solving quadratic equations, especially when the equation can be expressed in the form \((x-a)^2 = b\). When we apply this property, we take the square root of both sides of the equation, giving us two possible solutions, one positive and one negative. This is expressed mathematically as \(x-a = \pm \sqrt{b}\).

• The \(\pm\) symbol indicates the dual nature of the solutions. We have to consider both possibilities: one where the square root is positive, and another where it is negative.
• This method simplifies the process as it bypasses the need for complex algebraic factoring.

In our specific case, setting the equation \((x-2)^2 = 25\) to use the square root property gives us \(x-2 = \pm \sqrt{25}\). The trick is to remember that every square root operation naturally involves both a positive and a negative outcome. This dual solution approach is essential in fully solving the equation.

Radicals Simplification

Radicals can seem tricky at first, but understanding how to simplify them is crucial in solving quadratic equations effectively. A radical expresses the root of a number, with the square root being the most common
To simplify a radical like \(\sqrt{25}\):

• First, identify if the number under the radical is a perfect square. Numbers like 4, 9, 16, and 25 are perfect squares as they can be expressed as a number multiplied by itself: for instance, \(5 \times 5 = 25\).
• Recognize the square root: The square root of 25 is 5. This because squaring 5 gives 25. Hence, \(\sqrt{25} = 5\).

Simplifying radicals saves time and effort by making the numbers simpler to work with. Instead of dealing with \(\pm \sqrt{25}\) in our equation, it becomes \(\pm 5\), which is straightforward and eliminates extra calculations.

Quadratic Solutions

Once the equation is simplified using these properties, the next step is solving for the unknown, typically represented as \(x\). Quadratic solutions require isolating \(x\) in the final expression, which leads to the result.
In our example, we reach the equation \(x - 2 = \pm 5\). Here's how to obtain the final values for \(x\):

• Add 2 to both sides, resulting in \(x = 2 \pm 5\). The symbol \(\pm\) leads to two separate calculations:
• For the positive solution, solve \(x = 2 + 5\), which gives \(x = 7\).
• For the negative solution, solve \(x = 2 - 5\), resulting in \(x = -3\).

These steps find the two possible solutions to the quadratic equation. Thus, \(x = 7\) and \(x = -3\) are the solutions, showing the dual nature of solutions in quadratic equations using the square root property. This approach simplifies the solving process dramatically and ensures all possible solutions are considered.